Bourbaki

Proposition 7

Tag 08LF, permanent. Cite it as [Bourbaki, Tag 08LF].

Integration, chapter III, § 1. Measures on a locally compact space

For every set S\mathscr{S} of bounded subsets of K(X;C)\mathscr{K}(X;\mathbf{C}) that is a covering of K(X;C)\mathscr{K}(X;\mathbf{C}), the space MS(X;C)\mathscr{M}_{\mathscr{S}}(X;\mathbf{C}) is Hausdorff and quasi-complete.

This results from the fact that K(X;C)\mathscr{K}(X;\mathbf{C}) is barreled (TVS, III, §4, No. 2, Cor. 4 of Th. 1).

Examples of measures. — I. Atomic measures. Let X be a locally compact space, aa a point of X; the mapping f↦f(a)f\mapsto f(a) of K(X;C)\mathscr{K}(X;\mathbf{C}) into C\mathbf{C} obviously satisfies the condition (4) with MK=1M_K=1 for every compact subset K of X containing aa, hence is a measure on X, which is denoted by εa\varepsilon_a; it is called the Dirac measure at the point aa, or the measure defined by a unit mass placed at the point aa.

More generally, let α\alpha be a mapping of X into C\mathbf{C} such that, for every compact subset K of X,

∑x∈K∣α(x)∣<+∞. \sum_{x\in K}|\alpha(x)|<+\infty.
Then, for every function f∈K(X,K;C)f\in\mathscr{K}(X,K;\mathbf{C}), the sum
μ(f)=∑x∈Xα(x)f(x) \mu(f)=\sum_{x\in X}\alpha(x)f(x)
is defined, being equal to
∑x∈Kα(x)f(x); \sum_{x\in K}\alpha(x)f(x);
it is clear that μ\mu is a linear form on K(X;C)\mathscr{K}(X;\mathbf{C}) and that, for f∈K(X,K;C)f\in\mathscr{K}(X,K;\mathbf{C}),
∣μ(f)∣≤(∑x∈K∣α(x)∣)⋅∥f∥, |\mu(f)|\leq\left(\sum_{x\in K}|\alpha(x)|\right)\cdot\|f\|,
in other words the condition (4) is satisfied.

A measure μ\mu on X is said to be atomic if there exists a mapping α\alpha of X into C\mathbf{C} such that

∑x∈K∣α(x)∣<+∞ \sum_{x\in K}|\alpha(x)|<+\infty
for every compact subset K of X, and such that μ\mu is equal to the measure defined as above. If N is the set of x∈Xx\in X such that α(x)≠0\alpha(x)\neq0, the condition imposed on α\alpha implies that for every compact subset K of X, K∩NK\cap N is countable. One also says that μ\mu is defined by the masses α(x)\alpha(x) placed at the points x∈Nx\in N. If one assumes that N∩KN\cap K is finite for every compact set K⊂XK\subset X, then obviously
∑x∈K∣α(x)∣<+∞; \sum_{x\in K}|\alpha(x)|<+\infty;
it comes to the same to say that N is a closed and discrete subspace of X, because then every point of X has a compact neighborhood containing only a finite number of points of N, and conversely, if this is the case, then every compact subset of X can be covered by a finite number of such neighborhoods. When N is closed and discrete, every atomic measure defined by a function α\alpha such that α(x)=0\alpha(x)=0 on CN\mathbf C N is called a discrete measure on X (cf. §2, No. 5).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

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Cites

Transcribed from the printed volume. Not translated.